The permeance

Completing of the equivalent circuit. The permeance of the air-gaps determination. Determination of the steel magnetic potential drops, initial estimate magnetic flux through the air-gap. Results of the computation of electromagnet subcircuit parameters.

The course project consists of four parts. In the first part the so-called direct problem is solved and the magnetic-moving force is found. In the second part the winding is calculated and the pure MMF of the coil is found. The third part includes the so-called inverse problem, the magnetic flux is being determined. And the last, the fourth part includes defining of the actuation time of the electromagnet.

This course design is intended for getting skills on calculation of electromagnets. In this work next problems are solved:

- direct problem. Condition of the work of electromagnet: the armature is in open stage. Here such parameters are calculated: permeances of the air-gaps, magnetic fluxes, magnetic potential drops in metal and, as a result, magnetic motive force;

- calculation of winding. Here such parameters are calculated: diameter and cross-section of wire, resistance of wire, current, flowing in wire and precise value of magnetic motive force;

- inverse problem. Conditions of the work: armature is in closed stage. Here such parameters are calculated: permeances of air-gaps, magnetic flux, inductance of metal and magnetic intensity;

- calculation of dynamic characteristics of electromagnet. Here such characteristics are found: moving, pick up, actuation time and electromagnetic force for actuation of calculated electric apparatus.

Nowadays special computer programs are made for computing electromagnets and their characteristics. But the importance of this work is to understand and to learn the stages of finding one or another thing during calculation of contacting electric apparatus.

1.2. Completing of the equivalent circuit

The first step is to set up equivalent magnetic circuit without taking into account reluctance of iron subcircuits. The scheme is represented on the Figure 1.2.

The MMF that is produced by flowing current through the coil is drawing as the source of power. The magnetic resistances of operating and parasitic air gaps are drawn as resistors.

Rm1 is the magnetic resistance of the operating air-gap. Rm2 and Rm3 are the magnetic resistances of the parasitic air-gaps.

Figure 1.2. - Equivalent circuit

1.3. The permeance of the air-gaps determination

The permeance of the operating air-gap is determined according to the method of probable flux passes. The fluxes in this air-gap are shown on the Figure 1.3.

Figure 1.3.1. - Fluxes flowing between core and armature (1) and between yoke and armature (2)

The magnetic permeances of the operating air-gap by the method of probable fluxes can be calculated

where µ0 - is magnetic constant, or vacuum permeability, ;

дп - is the value of the operating air-gap, m.

where дп - is the value of the operating air-gap, m.

Now let's find the front magnetic flux permeance of the operating air-gap:

.

Now it is possible to find the full magnetic permeance of the operating air-gap:

,

.

The next step is to find the magnetic permeance of parasitic air-gap д2. It can be found using specific permeance for coaxial cylinders [5]. The flux through д2 can be calculated without taking into account fluxes of bulding, because the dimensions of the pole are more than the air-gap in 240 times.

where d2 - is the value of the parasitic air-gap, m.

.

Now let's find the permeance of the parasitic air-gap between the yoke and the armature using Roters' method.

At first, let's find square of a high yoke surface:

;

The front magnetic flux permeance of the parasitic air-gap is determined:

;

.

For Лд1z1 the formula of half-ring is used [5]:

,

For Лд1p1 the formula of half-cylinder is used [5]:

For Лд1p2 the formula of quarter-cylinder is used [5]:

,

For Лд1z2 the formula of quarter-ring is used [5]

,

The total permeance of the parasitic air-gap is found:

1.4. Determination of magnetic flux

Now it is necessary to find the flux of the pole:

where Bд - is magnetic induction, T;

Sд - is the pole face area, m2;

увп - is the coefficient of bulging.

The bulging coefficient is to be found:

,

where Лд - is the permeance of the operating air-gap, H;

ЛдT - is the front magnetic flux permeance of the operating air-gap, H.

Then the pole face area is calculated

;

;

where Bд - is magnetic induction, T;

увп - is the coefficient of bulging;

d- is the diameter of the core, m.

1.5. Determination of the steel magnetic potential drops

Now let's find the magnetic potential drops of the magnetic circuit by the method of leakage ratio. The division of the electromagnet is shown on the Figure

1.5.1 At first the core is divided by 5 equal parts (x) equaled to

Figure 1.5.1. - Dividing of a magnetic circuit on subcircuits

Now the leakage ratios of every subcircuit must be defined:

where xi - is the length from the top of the core to the middle of the investigated subcircuit;

лS - is the specific permeance; in this case this is the specific permeance of coaxial cylinders:

.

After that, the total permeance ЛдУ is found:

;

;

,

where Лд, Лд1, Лд2 - are the permeances of the air-gaps, H.

Then, the calculations for the first subcircuit are made. The bulging coefficient уX1 is found

,

where лS - is the specific permeance, ; in this case this is the specific permeance of coaxial cylinders;

ЛдУ - is the total permeance, H.

Then the magnetic flux through the first subcircuit is made:

;

,

where Фд - is the magnetic flux through the circuit, Wb;

уX1 - is the bulging coefficient of the subcircuit.

After it, the induction of the core is found:

,

where ФX1 - is the magnetic flux through the subcircuit, Wb;

d - is the diameter of the core, m.

The induction of the yoke is found:

;

.

From the graph of magnetic curves [5] the magnetic intensity is taken:

;

.

Now let's find magnetic potential drop of the first subcircuit:

;

.

Calculations for the next subcircuits are done by analogy to the first one.

Now the magnetic potential drop in the armature can be found. At first, the magnetic flux through the armature is found:

;

After it, the induction of the armature is found:

From the graph of magnetic curves [5] the magnetic intensity is taken:

.

Now let's find magnetic potential drop:

;

.

Now the magnetic potential drop in the base of the electromagnet can be found. The parasitic air-gap can be neglected and the base can be decided to be entire, because the existence of such a little air-gap almost doesn't influence on the flux. The bulging coefficient уX6 is found:

,

where лS - is the specific permeance, ; in this case this is the specific permeance of coaxial cylinders;

ЛдУ - is the total permeance, H.

Then the magnetic flux through the base is found:

;

;

After it, the induction of the base is found:

;

,

where Фbase - is the magnetic flux through the base, Wb.

From the graph of magnetic curves [5] the magnetic intensity is taken:

Now let's find magnetic potential drop:

;

,

where Hbase - is the magnetic intensity of the base, .

1.6. Determination of the magnetic motive force of the electromagnet

Now let's find the MMF

where Umi - are magnetic potential drops, A;

Фi - is the magnetic flux, Wb;

Лдi - is the permeance, H.

1.7. Results of the computation of electromagnet subcircuit parameters

Table 1.7.1. - Parameters of the electromagnet subcircuits

уi

Фi, mWb

Bi, T

Hi, A/m

Umi, A

Фi/Лi, A

Дxi, m

xi, m

xy1

1.69

0.206

0.011

1

0.05

-

0.05

0.025

xy2

2.81

0.343

0.046

2

0.1

-

0.05

0.075

xy3

3.66

0.447

0.073

3

0.15

-

0.05

0.125

xy4

4.23

0.516

0.084

3.5

0.175

-

0.05

0.175

xy5

4.51

0.55

0.09

3.9

0.195

-

0.05

0.225

xc1

1.69

0.206

0.42

14

0.7

-

0.05

0.025

xc2

2.81

0.343

0.67

70

3.5

-

0.05

0.075

xc3

3.66

0.447

0.91

400

20

-

0.05

0.125

xc4

4.23

0.516

0.98

700

55

-

0.05

0.025

xc5

4.51

0.55

1.01

1500

75

-

0.05

0.075

xarm

1

0.124

0.08

3

0.32

-

0.1025

0.125

xbase

4.54

0.554

0.28

8

1.64

-

0.1025

0.25

д

1

0.124

-

-

-

193

0.01

-

д1

1

0.124

-

-

-

97

0.01

-

д2

4.54

0.554

-

-

-

76

0.00025

-

2. CALCULATION OF WINDING

2.1. Determination of final MMF of a winding, taking into account safety factor

The task is to calculate the winding knowing next parameters: l=250 mm; h=80 mm; (Iw)=1521.64 A; U=36 V. Winding is made of copper. б0=4.3•10-3 K-1 is the temperature coefficient of copper; с0=1.62•10-8 ?•m - is specific resistance of copper; q=105°C - is permissible heat temperature of the winding. I determine finally MMF of a winding taking into consideration safety factor kЗ taken from the interval 1.3…1.5.

,

where (Iw) - is the magnetic motive force of the electromagnet A;

2.2. Calculation of the winding parameters

The resistance of copper сq can be defined:

,

where с0=1.62•10-8 ?•m - is specific resistance of copper;

б0=4.3•10-3 K-1 - is the temperature coefficient of copper;

q=105°C - is permissible heat temperature of the winding.

Average length of one turn

,

where h - is the distance between the core and the yoke, m;

d - is the diameter of the core, m.

but keeping some place in reserve, let

,

where (Iw)final - is the magnetic motive force of the electromagnet A;

сq - is the resistance of copper, ?•m;

lв.ср. - is average length of one turn, m;

U - is voltage, V.

Now one must choose the most suitable diameter of the wire;

Then the cross-section of the wire is to be found:

,

where dw - is the diameter of the wire, m.

Then I find the number of turns in the coil

,

where U - is voltage, V;

jдоп - is current density, A/m2;

сq - is the resistance of copper, ?•m;

lв.ср. - is average length of one turn, m.

Then the square of the winding is to be found;

,

where w - is the number of turns in the coil;

q0 - is the cross-section of the wire;

kз - is the space factor coefficient equaled to 0.656.

Then the width of the winding can be found:

,

where Q0 - is the square of the winding, mm2;

l - is the length of the yoke, mm.

equivalent circuit magnetic steel

2.3. Determination of pure resistance and MMF of the wire

At first, I find the resistance of the coil:

,

where сq - is the resistance of copper, ?•m;

w - is the number of turns in the coil;

lв.ср. - is average length of one turn, m.

q0 - is the cross-section of the wire, m2.

Then it is possible to calculate the current flowing in the winding:

,

where U - is the voltage in the winding, V;

R - is the resistance of the coil.

Now it is necessary to check whether the current density was chosen correctly:

He bulging fluxes will be neglected as the dimensions of the pole. According to equivalent circuit, figure 2.1., operating and parasitic air-gaps permeances will be determined.

Operating air-gap:

,

where µ0 - is magnetic constant, or vacuum permeability, ;

d - is the diameter of the core, m;

дп - is the value of the operating air-gap, m.

Parasitic air-gaps:

,

where µ0 - is magnetic constant, or vacuum permeability, ;

h - is the distance between the core and the yoke, m;

a - is the thickness of the core, m;

дп - is the value of the operating air-gap, m;

d - is the diameter of the core, m.

- it was found in the 1st part.

Total permeance is equal to:

3.2. Determination of initial estimate magnetic flux through the air-gap

According to the determination of MMF:

where Ф - is the magnetic flux, Wb;

Лд - is the magnetic permeance, H.

In our case we may use for first approximation the next formula:

from which:

Then the induction of the core is found:

where Ф - is the magnetic flux, Wb;

Лд - is the magnetic permeance, H;

d - is the diameter of the core, m.

As the found induction is greater than the saturation induction the saturation one is taken from the graph of magnetic curves:

,then

From this graph we also must find magnetic intensity of the steel:

Then the induction of the armature and the base of the yoke is found

,

where Ф - is the magnetic flux, Wb;

D - is the diameter of the yoke, m;

a - is the thickness of the core, m.

From the graph of magnetic curves the magnetic intensity of steel is found:

Then one can find the induction of the yoke:

,

where Ф - is the magnetic flux, Wb;

D - is the diameter of the yoke, m;

h - is the distance between the core and the yoke, m.

From the graph of magnetic curves the magnetic intensity is found:

Now let's find the potential magnetic drop of the iron:

Then, the magnetic flux can be found:

After that everything from the formula 3.1 should be repeated till the relative difference between previous and new approximation of magnetic flux is equal to 5%.

1

2

3

4

Ф, mWb

0.54

0.43

0.405

0.398

Bc, T

1.1

0.88

0.78

0.76

Ba/Bb, T

0.35

0.24

0.18

0.15

By, T

0.09

0.077

0.059

0.033

Hc, A/m

4000

500

100

14

Ha/Hb, A/m

10

9

8

7

Hy, A/m

3

2.9

2.3

2

3.3. Determination of electromagnet force

As д=0.1 mm, then Maxwell's formula will be used for finding electromagnetic pulling force:

,

where Ф - is the magnetic flux, Wb;

µ0 - is magnetic constant, or vacuum permeability, ;

S - is the square of the core, m2.

4. Defining of actuation time

To calculate actuation time, knowing next parameters:

Then the average magnitude of counteracting force is to be found:

Then the inductance of the winding is to be found:

,

where ЛдУ - is the total permeance of the electromagnet, H;

w - is the number of turns in the coil.

Then the pick-up time of the electromagnet is to be found:

,

where L - is the inductance of winding, H;

R - is the resistance of the coil, ?;

kз - is the assurance factor.

Now let's find the mass of movable parts:

,

where V - is the volume of the armature, m3;

сst - is the density of steel, kg/m3.

Then the moving time is determined:

,

where m - is the mass of the movable parts, kg;

дп - is the value of the operating air-gap.

Actuation time is equal to:

.

Conclusions

An electromagnet is simply a coil of wire. It is usually wound around an iron core. However, it could be wound around an air core, in which case it is called a solenoid. When connected to a DC voltage or current source, the electromagnet becomes energized, creating a magnetic field just like a permanent magnet. The magnetic flux density is proportional to the magnitude of the current flowing in the wire of the electromagnet. The polarity of the electromagnet is determined by the direction the current. The north pole of the electromagnet is determined by using your right hand. Wrap your fingers around the coil in the same direction as the current is flowing (conventional current flows from + to -). The direction your thumb is pointing is the direction of the magnetic field, so north would come out of the electromagnet in the direction of your thumb. DC electromagnets are principally used to pick up or hold objects.

When connected to an AC voltage or current source, the electromagnet will be changing its flux density as the current fluctuates. The polarity of the magnet will also change as the current reverses direction every half cycle. AC electromagnets can be used to demagnetize objects (like TV screens, audio tapes, vcr tapes) or to hold objects. However, due to the inductance of the electromagnet, the AC current that will actually flow will be reduced when compared to a DC voltage equal to the RMS value of the AC voltage feeding the electromagnet.

The key importance of an electromagnet is the ability to control the strength of the magnetic flux density, the polarity of the field, and the shape of the field. The strength of the magnetic flux density is controlled by the magnitude of the current flowing in the coil, the polarity of the field is determined by the direction of the current flow, and the shape of the field is determined by the shape of the iron core around which the coil is wound.

The electromagnets are used in lots of things. Motors are the most prominent example. Every electric motor uses at least one electromagnet. Most use two, one stationary and one moving. Sometimes one or the other of them is a permanent magnet. Generators, being the alter ego of motors, also use electromagnets. Loudspeakers and earphones use electromagnets to drive the diaphragm. Televisions use electromagnets to direct the electron beam on the screen. Scrap heaps use electromagnets to pick up large ferrous items like junked cars. Also fire doors as they can shut after detecting a fire the doors lock after a few minutes so people are safe outside and don't return back into the building.

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